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Se dau trei matrice de ordin 3 cu elemente numere reale, \( A,B,C \) astfel incat \( \det(A)=\det(B)=\det(C) \) si \( \det(A+iB)=\det(B+iC) \). Aratati ca \( \det(A+B)=\det(A+C) \).

\( \{A,B,C\}\subset\mathrm M_3(R)\ \wedge\ \det A=\det B=\det C\ \wedge\ \det (A+iB)=\det (C+iA) \) \( \Longrightarrow \) \( \det (\mathrm A+\mathrm B)=\det (\mathrm C+\mathrm A)\ . \)

Dem. Notam \( f(x)=\det (\mathrm{A+xB})-\det (\mathrm{C+xA})\ ,\ x\in\mathbb R\ . \) Se observa ca \( f(0)=\underline{\det A-\det C=0}\ \ ,\ \underline{f(i)=0}\ . \)

Polinomul \( f \) are coeficienti reali si gradul \( \mathrm{gr} f\le 3\ . \) Asadar \( f(x)=kx\left(x^2+1\right)\ . \) Deci \( k=\lim_{x\to\infty}\ \frac {f(x)}{x^3}= \)

\( \lim_{x\to\infty}\ \frac {1}{x^3}\cdot \left[\det (\mathrm{A+xB})-\det (\mathrm{C+xA})\right]=\lim_{x\to\infty}\ \left[\det \left(\mathrm{\frac 1x A+B}\right)-\det \left(\mathrm{\frac 1xC+A}\right)\right]=\underline{\det \mathrm B-\det \mathrm A=0}\ . \)

In concluzie \( k=0 \) , adica \( f \) este polinomul nul. In particular, \( f(1)=0 \) , adica \( \det (\mathrm A+\mathrm B)=\det (\mathrm C+\mathrm A)\ . \)

Remarca. Am folosit faptul ca \( a\in\mathbb R\ \wedge\ A\in\mathrm M_n(R) \) \( \Longrightarrow \) \( \det\left(\mathrm{a\cdot A}\right)=a^n\cdot \det \mathrm A\ . \)

Notam \( P=\det(A+xB)=\det(B)x^3+ax^2+bx+\det(A), Q=\det(C+xA)=\det(A)x^3+cx^2+dx+\det(C) \) doua polinoame. Stim ca \( P(i)=Q(i) \). Deoarece matricile sunt reale, si coeficientii \( a,b,c,d \) sunt reali. Folosind faptul ca determinantii matricelor sunt egali obtinem ca \( -a+bi=-c+di \) adica \( a=c,b=d \) si polinoamele coincid, adica \( P(1)=Q(1) \), ceea ce trebuia demonstrat.